CHAPTER 1 1 INTRODUCTION TO Prepared By Sheilani binti Shaari Mahizan binti Mustapha Department of...

CHAPTER 1

INTRODUCTION TO

Prepared By

Sheilani binti Shaari

Mahizan binti Mustapha

Department of Electrical Engineering/PKB

INTRODUCTION TO

ELECTRIC CIRCUIT

Upon completion of this course, students

Course Learning Outcome (CLO)

Department of Electrical Engineering PKB

Upon completion of this course, students

should be able to:

Explain clearly basic electrical quantities,

types of electrical circuits, electrical power,

electrical energy and solve related

problems.

ELECTRICAL


ELECTRICAL QUANTITIES

INTRODUCTION


ELECTRIC CIRCUIT

OHM’S LAW

CONTENT


OHM’S LAW

ELECTRICAL POWER

ELECTRIC ENERGY

General Objective (GO)

Student should be able to:

1. Understand basic electrical quantities

LEARNING OUTCOMES


quantities

INTRODUCTION

Electric is an energy which cannot

seen but can be felt and be used by

human on today and future.


Electric energy can be created impact from action:

a) Friction

b) Heat

c) Electromagnetic field

Electric energy can be change to another some energy like:

a) Heat energy – Iron

INTRODUCTION


a) Heat energy – Iron

b) Sound energy – Radio

c) Kinetic Energy – Motor

d) Light energy – Lamp

2 types of electric:

i) Static electricity

no electron movement in certain

INTRODUCTION


no electron movement in certain direction.

ii) Dynamic electricity

electron movement in certain direction.

Item Symbol Unit

Electromotif Force E Volt (V)

Electrical Charge Q Coulomb (C)



Electrical Charge Q Coulomb (C)

Current I Ampiar (A)

Voltage V Volt (V)

Resistance R Ohm (Ω)

Resistivity ρ Rho (Ωm)

Electromotive Force (Emf)The force or electric pressure that causes the flow of electrons or current in the circuit



current in the circuit

Battery

Electrical Charges

The Positive and negative charges



charges

+ve -ve

Current

The movement of the electrical charge cause by free electron movement from



electron movement from positive to negative terminals.

R

V

-ve+ve

Voltage

The Potential difference between 2 points.



between 2 points.

+ve -ve

V

Resistance

The property of material which is opposition to current flow



RI

R

I

Resistance

4 factor that influence the value of resistance l



1. Length of conductor,

2. Surface area , A

3. Resistivity, ρ

4. Conductor Temperature, T

Resistance1. Length of conductor,

The length of conductor is proportional to the resistance. The longer the length of the wire, the higher the resistance value.



higher the resistance value.

2. Surface area , A

Area is inverse proportional to the resistance. As the resistance increase the cross section area of a conductor will decreases.

l∝R

AR

1∝

Resistance3. Resistivity

Resistivity is proportional to the resistances. Higher the resistance, higher the resistivity.



4. Conductor Temperature, T

The conductor temperature is proportional to the resistance. As the conductor temperature increase the value of resistance also increase.

ρ∝R

TR ∝

Resistivity, ρThe characteristic of conductive material to opposition or decrease the current flow



the current flow

R=ρℓ/A

R=ρℓ/A

Where;

A = ∏(r)2

A = ∏(d/2)2



Where;

A= cross section area (m)

ρ= Resistivity (Ωm)

ℓ = Length (m)

R= Resistance

1. Calculate the resistance of a

1.5m length of alumuinium wire. Given diameter wire 10mm and

R?Example



Given diameter wire 10mm and resistivity of alumuinium is 0.025µΩm

ℓ

dρ

Item Spesification Material

conductor Material that current is easily flow

Copper

iron

Semi Material that has Silicon



Semi conductor

Material that has conductance value between conductor and insulator

Silicon

germinium

Insulator Material that does not allowed current flow in normal condition

Rubber, glass, air

SOLVE PROBLEMS


SOLVE PROBLEMS(Q1 – Q11)

…CONTINUED

ELECTRICAL


ELECTRICAL CIRCUIT

INTRODUCTION


ELECTRIC CIRCUIT

OHM’S LAW

CONTENT


OHM’S LAW

ELECTRICAL POWER

ELECTRIC ENERGY


Student should be able to :

1. Explain types of electrical circuit

LEARNING OUTCOMES


2. Understand Ohms Law

3. Understand Electrical Power

4. Understand Electrical Energy

Electric circuit is a combination or cable to bring current from voltage sources to electrical components (load)

ELECTRIC CIRCUIT


(load)

It consists of two type;

i) Complete electric circuit

ii) Non Complete electric circuit

Complete electric circuit

Also called basic/

simple circuit

it is closed end

I

V

R

ELECTRIC CIRCUIT


it is closed end connection that can make current through completely.

The circuit must have voltage supply (V), Electric current (I) and resistance (R).

Non Complete electric circuit

It is a circuit without one of three components either voltage sources or load resistance

ELECTRIC CIRCUIT


or load resistance

The current flow will never happen with perfect in non complete circuit.

2 type of Non Complete electric circuit;

i) Open circuit

ELECTRIC CIRCUIT


Open circuit

ii) Short circuit


i. Open circuit

The load in that circuit will open

No current flow occur

ELECTRIC CIRCUIT


No current flow occur

Resistance value is higher (∞)

Load (R)Pull outV


ii.Short circuit

Connection at the load will short with a conductor which no resistance.

ELECTRIC CIRCUIT


a conductor which no resistance.

The current which go through is high.

Fuse will burntV

R

Short withcable

Definition

The current flow through the electrical circuit is

OHM’S LAW


electrical circuit is directly proportional to the potential difference across the circuit and inversely proportional to the resistance of the circuit.

l∝RI V

OHM’S LAW

The electric current (I) is :

directly proportional to the voltage


AR

1∝

l∝RI V

IR

inversely proportional to the resistance.

OHM’S LAW

IRV =


V = IR

Voltage (V) = Current (A) x Resistance ( Ω)

IRV =I = V

R

R = V

I

Linear Resistance

With Constant Value of Resistance:

Increasing Voltage will increasing

OHM’S LAW


Increasing Voltage will increasing the current

l∝RV IR CONSTANT

Linear Resistance

Constant Resistance and temperatureV

OHM’S LAW


temperatureV

I

R constant

Graph V vs I with constant R

Non Linear Resistance

Changing Value of R

Increasing Voltage will increasing

OHM’S LAW


Increasing Voltage will increasing the current to a constant level

l∝RV IR CHANGING

Non Linear Resistance

Changing value of R

V

OHM’S LAW


V

I

R changing

Graph V vs I with changing value of R

Constant level

increasing

Calculate the current flow in the circuit ifthe resistance is 10Ω and the supply

I?

R

Example

OHM’S LAW


the resistance is 10Ω and the supplyvoltage is 15V.

Then, calculate the current value, if thevalue of resistance is change to 10 kΩ.

R

V

I?

R

ELECTRIC POWER

is a job that can be done in one time unit

IVP =


Power (P) = Current (A) x Voltage (V)

Unit : Watt (W)

IVP =

Power (P) = Current 2 (A) x Resistance (R)

RIP2

=

ELECTRIC POWER


R

VP

2

=

Power (P) = Voltage 2 (v) / Resistance (R)

WATT METER

used to measure the value of the power that has been used

ELECTRIC POWER


W

Internal Connection

Meter Symbol

WATT METERThere are two coils in it:

The CURRENT COIL is connected in series with the circuit.

ELECTRIC POWER


VS

LoadVOLTAGE COIL connected in parallel

with the circuit.

ELECTRIC ENERGY

Electrical energy is the product of power and time.

T = Pt


T = PtEnergy (T) = Power (P) x Time (hour)

Unit : Kilowatt hour (kWh) or Joule (watt second)

T = VItEnergy (T) = Voltage (V) x Current (I) x Time (hour)

T = I2Rt

ELECTRIC ENERGY


T = I2RtEnergy (T) = Current2 (I) x Resistance (R) x Time (hour)

T = (V2/R)tEnergy (T) = [Voltage2 (V) / Resistance (R) ] Time (hour)

is the energy absorbed to supply load of 1 KW for a period of 1 hour.

Work (w) = Power (P) x Time (h)

WORK


Work (w) = Power (P) x Time (h)

Watt

is the power used when 1A current flow between two points with a 1 volt potential

Unit for work or energy is Joule.

1 Joule = 1 watt second

WORK


1 Joule = 1 watt secondWork (J) = Power (P) x Time (s)

Notes: Convert hour to seconds

1 H = 1 x 60 minutes x 60 sec

= 3600 s

Example

A toaster used 5A current and 240 V

I = 5A

ELECTRIC ENERGY


A toaster used 5A current and 240 V supply for 15 minutes. Calculate,

i. Power (P) usedii. Energy (T) which absorbed the

heat in kJ

v = 240vt = 15 min

SOLVE PROBLEMS



…CONTINUED

ELECTRICAL


ELECTRICAL CIRCUIT ANALYSIS

INTRODUCTION

RESISTOR CIRCUIT ANALYSIS

KIRCHOFF’S LAW

CONTENT



Student should be able to:

1. Understand the characteristics of series and parallel circuits.

LEARNING OUTCOMES


series and parallel circuits.

2. Explain Kirchhoff’s Law in complex electrical circuits.

Electrical circuit

Is a complete connection to the load by using a conductors

INTRODUCTION


The load or resistor in the circuit can be connected in 3 different ways:

INTRODUCTION


1. Series Circuit

2. Parallel Circuit

3. Combination of Series and Parallel Circuits

refer to the connection of the resistor in the circuit.

connected from end to end.

SERIES CIRCUIT


VT

R1 R2 R3

V2 V3V1

Total Resistance RT

Is the sum of all the resistor that exist in the circuit

SERIES CIRCUIT


RT = R1+R2+R3

Total Current ITThe current through each resistor is equal to the total current .

I1

SERIES CIRCUIT


IT=I1=I2=I3

ITI2

I3

Total voltage

is the sum of all voltage drop (voltage down) on each of the resistance

SERIES CIRCUIT


VT=V1+V2+V3V1

V2

V3

Voltage drop

is the reduction of the voltage supply in every resistance

V2 = I2 R2

SERIES CIRCUIT


V1

V2

V3

V1 = I1 R1

V2 = I2 R2

V3 = I3R3

voltage divider law

is used to determine voltage value which across every resistance in series circuit

R

SERIES CIRCUIT


VT

R1

R2

TVRR

RV )(

21

1

1+

=

TVRR

RV )(

21

2

2+

=

V1

V2

voltage divider law

V1R1

TVRRR

RV )(

321

11

++=

SERIES CIRCUIT


VT

V2R2

R3 V3

TVRRR

RV )(

321

22

++=

TVRRR

RV )(

321

3

3++

=

Example

Based on the circuit below. Calculate:

a. Total resistance, RT

b. Total Current, IT

SERIES CIRCUIT


b. Total Current, ITc. Voltage drop across each resistor by

using:

i. Ohm’s Law

ii. Voltage Divider Law

VT = 120 v

R1 = 15Ω

R2 = 10Ω

Solution

a. Total resistance RT = R1 + R2

RT = 15 + 10

RT = 25 Ω

R1 = 15Ω

SERIES CIRCUIT


RT = 25 Ω

b. Total Current IT = VT / RT

IT = 120/25

IT = 4.8 A

Notes: Series Circuit : IT = I1 = I2

VT = 120 vR2 = 10Ω

Solution

c. Voltage drop across each resistor by using Ohm’s Law

V1 = I1 R1

= 4.8 x 15

SERIES CIRCUIT


Notes: Series Circuit : V1 + V2 = VT

VT = 120 v

R1 = 15Ω

R2 = 10Ω

V1

V2

1 1 1

= 4.8 x 15= 72 v

V2 = I2 R2

= 4.8 x 10= 48 v

Solution

c. Voltage drop across each resistor by using Voltage Divider Law

R = 15Ω R=

SERIES CIRCUIT


VT = 120 v

R1 = 15Ω

R2 = 10Ω

V1

V2

= 72 v

= 48 v

TVRR

RV )(

21

11

+=

TVRR

RV )(

21

22

+=

120)1015

10(2

+=V

120)1015

15(1

+=V

Notes:

The answer for V1 and V2 is same for both Laws

connection of resistors which is against between each other

PARALLEL CIRCUIT


VTR1 R2

Total resistance

PARALLEL CIRCUIT


VTR1 R2

R3

321

1111

RRRRT

++=

Total voltage

voltage across each parallel resistor is equal to the voltage source

PARALLEL CIRCUIT


VT V2 V3

321 VVVVT ===

V1 V2 V3

Total current

the summation of all current flow in each branch is equal to the current source.

I I

PARALLEL CIRCUIT


VT

I1 I2 I3IT

321 IIII T ++=

Current divider law

is used to determine the value of the current in each branch

PARALLEL CIRCUIT


Ij

R1 R2V1

V2

I1I2

VT

TIRR

RI )(

21

2

1+

=

jIRR

RI )(

21

12

+=

Example


1. Total resistance, RT

2. Voltage drop across each resistor

PARALLEL CIRCUIT



3. Current I1 and I2 by using:

Ohm’s Law

Current Divider Law

Vj = 240v

IT

I1 I2

R1 = 2Ω R2 = 4Ω

Solution

1. Total resistance

1/RT = 1/R1 + 1/R2

1/RT = 1/2 + ¼

PARALLEL CIRCUIT


1/RT = 1/2 + ¼

1/RT = ¾

RT = 4/3

RT = 1.33 Ω

VT = 240v

IT

I1 I2

R1 = 2Ω R2 = 4Ω

Solution


IT

PARALLEL CIRCUIT


VT= 240v

IT

R1 = 2Ω R2 = 4ΩV1 V2

Parallel circuit:

V1 = V2 = VT = 240 v

Solution

3. Current by using Ohm’s Law

I1 = V1 /R1

= 240/2

IT = I1 + I2= 120 + 60

PARALLEL CIRCUIT


Vj = 240v

IT

I1I2

R1 = 2Ω R2 = 4Ω

1 1 1

= 240/2= 120 A

I2 = V2 /R2

= 240/4= 60 A

= 120 + 60= 180 A

Solution

3. Current by using Current Divider Law

IR

I )( 2=

PARALLEL CIRCUIT


VT= 240v

IT

R1 = 2Ω R2 = 4Ω

I1

= 120 A

TIRR

I )(21

2

1+

=

180)42

4(1

+=I

Solution

3. Current by using Current Divider Law

IR

I )( 1=

PARALLEL CIRCUIT


VT= 240v

IT

R1 = 2Ω R2 = 4Ω

I2

= 60 A

TIRR

RI )(

21

12

+=

180)42

2(2

+=I

Notes:

The answer for I1 and I2 is same for both Laws

Most of the electric circuit are the combination of series and parallel circuits.

COMBINATION CIRCUIT


R1 R2

R3

IT

I1

I2I3

+ -

VT

to solve problems in calculating the value of total resistance, total voltage and total current for combinations circuit have to used both formulas in series and parallel circuit respectively.

COMBINATION CIRCUIT


parallel circuit respectively.

Example


1. Total resistance, RT

2. Current IT, I1 and I2

COMBINATION CIRCUIT


2. Current IT, I1 and I23. Voltage drop across

each resistor

R1 =10Ω R2 =20Ω

R3 15Ω

IT

I1

I2I3

+ -

VT =120v

used to solve complex electrical circuit which have 2 or more supply

There are 2 types of Kirchoff’s law;

KIRCHOFF’S LAW


There are 2 types of Kirchoff’s law;

a) Kirchoff’s current law

b) Kirchoff’s voltage law

Kirchoff’s current law(first order of Kirchoff’s law)

sum of the current into a node is equal to the sum of current out of the node

algebraic sum of all the current entering and leaving a

KIRCHOFF’S LAW


algebraic sum of all the current entering and leaving a node is equal.

I1321 III +=I3

I2

NODE

Kirchoff’s voltage law(second order of kirchoff’s law)

sum of the voltage drop and voltage source around the closed path is equal to zero

sum of voltage drop must equal to voltage source

KIRCHOFF’S LAW


sum of voltage drop must equal to voltage source

321 VVVVs ++=Vs

V1

V2

V3

Example

By using Kirchoff’s Law, calculate the current at each branch

KIRCHOFF’S LAW

IKirchoff’s Current Law


5v10v

R1 = 1Ω R2 = 6Ω R3 = 2Ω

I1

I3

I2

Kirchoff’s Current Law

Kirchoff’s Voltage Law

C1

Kirchoff’s Voltage Law

C2

SOLVE PROBLEMS



NEXT:

CHAPTER 2

INDUCTOR,


INDUCTOR, CAPACITOR AND ALTERNATING CURRENT

CHAPTER 1 1 INTRODUCTION TO Prepared By Sheilani binti Shaari Mahizan binti Mustapha Department of...

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